ABCD is a trapezium such that `AB and CD` are parallel and `BC _|_ CD`. If /_ADB= theta , BC =p and CD=q`, then AB is equal to

Options

A. `(p^2+q^2)/(p^2cos theta+q^2sintheta)`
B. `((p^2+q^2)sin theta)/(pcos theta+qsintheta)^2`
C. `((p^2+q^2)sin theta)/(pcos theta+qsintheta)`
D. `(p^2+q^2sin theta)/(pcos theta+qsintheta)`

Answer

To solve the problem, we need to find the length of side \( AB \) in trapezium \( ABCD \) where \( AB \parallel CD \) and \( BC \perp CD \). Given that \( \angle ADB = \theta \), \( BC = p \), and \( CD = q \), we can follow these steps:

### Step-by-step Solution:

1. **Identify the trapezium and angles**:
- We have trapezium \( ABCD \) with \( AB \parallel CD \) and \( BC \perp CD \).
- Given \( \angle ADB = \theta \).

2. **Label the sides**:
- Let \( BC = p \) and \( CD = q \).
- Since \( BC \perp CD \), \( BC \) acts as a height of the trapezium.

3. **Determine angles**:
- Let \( \angle ABD = \alpha \). By the properties of parallel lines and transversals, \( \angle ABD = \angle BDC = \alpha \).
- Therefore, \( \angle DAB = 180^\circ - \theta - \alpha \).

4. **Use the right triangle \( BCD \)**:
- In triangle \( BCD \), we can apply the Pythagorean theorem:
\[ BD^2 = BC^2 + CD^2 = p^2 + q^2 \]
- Thus, \( BD = \sqrt{p^2 + q^2} \).

5. **Relate angles using trigonometric ratios**:
- From triangle \( ABD \):
\[ \tan(\alpha) = \frac{BC}{CD} = \frac{p}{q} \]
- Therefore, we can express \( \tan(\alpha) \) as:
\[ \tan(\alpha) = \frac{p}{q} \]

6. **Apply the sine rule in triangle \( ABD \)**:
- According to the sine rule:
\[ \frac{AB}{\sin(\theta)} = \frac{BD}{\sin(180^\circ - \theta - \alpha)} \]
- Since \( \sin(180^\circ - x) = \sin(x) \):
\[ \frac{AB}{\sin(\theta)} = \frac{BD}{\sin(\theta + \alpha)} \]

7. **Substituting \( BD \)**:
- We know \( BD = \sqrt{p^2 + q^2} \):
\[ AB = \frac{\sqrt{p^2 + q^2} \cdot \sin(\theta)}{\sin(\theta + \alpha)} \]

8. **Using the sine addition formula**:
- The sine addition formula gives us:
\[ \sin(\theta + \alpha) = \sin(\theta) \cos(\alpha) + \cos(\theta) \sin(\alpha) \]
- Substituting this into our equation for \( AB \):
\[ AB = \frac{\sqrt{p^2 + q^2} \cdot \sin(\theta)}{\sin(\theta) \cos(\alpha) + \cos(\theta) \sin(\alpha)} \]

9. **Finding \( \cos(\alpha) \) and \( \sin(\alpha) \)**:
- From triangle \( BCD \):
- \( \cos(\alpha) = \frac{q}{\sqrt{p^2 + q^2}} \)
- \( \sin(\alpha) = \frac{p}{\sqrt{p^2 + q^2}} \)

10. **Final expression for \( AB \)**:
- Substituting \( \cos(\alpha) \) and \( \sin(\alpha) \) back into the equation for \( AB \):
\[ AB = \frac{\sqrt{p^2 + q^2} \cdot \sin(\theta)}{\sin(\theta) \cdot \frac{q}{\sqrt{p^2 + q^2}} + \cos(\theta) \cdot \frac{p}{\sqrt{p^2 + q^2}}} \]
- Simplifying gives:
\[ AB = \frac{(p^2 + q^2) \sin(\theta)}{q \sin(\theta) + p \cos(\theta)} \]

### Final Answer:
\[ AB = \frac{(p^2 + q^2) \sin(\theta)}{q \sin(\theta) + p \cos(\theta)} \]